JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
\(5.1\, g\, NH_4SH\) is introduced in \(3.0\, L\) evacuated flask at \(327\,^oC\). \(30\%\) of the solid \(NH_4SH\) decomposed to \(NH_3\) and \(H_2S\) as gases. The \(K_p\) of the reaction at \(327\,^oC\) is (\(R = 0.082\, L\, atm\, mol^{-1}\, K^{-1}\), Molar mass of \(S = 32\, g\, mol^{-1}\), molar mass of \(N = 14\, g\, mol^{-1}\))
- A \(0.242\times10^{-4}\, atm^2\)
- B \(1\times10^{-4}\, atm^2\)
- C \(4.9\times10^{-3}\, atm^2\)
- D \(0.242\,atm^2\)
Answer & Solution
Correct Answer
(D) \(0.242\,atm^2\)
Step-by-step Solution
Detailed explanation
\mathop {\mathop {\mathop {N{H_4}HS(s)}\limits_{5.1\,g} }\limits_{0.1\,g\,mol\, - 0.03} }\limits_{V = 3L,\,T = 327{\,^o}C} \leftrightarrow \mathop {\mathop {\mathop {N{H_3}(g)}\limits_{} }\limits_{0.03\,mol} }\limits_{\frac{{0.98}}{2}} + \mathop {\mathop {\mathop…
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